# Thread: Calculate volume between two curves (on left side of y axis) rotate about the y-axis

1. ## Calculate volume between two curves (on left side of y axis) rotate about the y-axis

What is the volume of the solid lying between the curves y= x^2+2x-4 and g=2x-3 ( on the left side of the y-axis) and is rotated about y-axis.

3b.jpg

2. Your drawing is good.
Your integral is almost good. $\pi$ seems to be missing.

Now what?

3. Originally Posted by tennisbossen
What is the volume of the solid lying between the curves y= x^2+2x-4 and g=2x-3 ( on the left side of the y-axis) and is rotated about y-axis.

3b.jpg
I presume the curves are supposed to be y=x^2+2x-4 and y=2x-3. But your work shows a function called f; what is that?

If you continue the way you are starting, f will have to be defined piecewise, so that your volume will be a sum of two integrals. That's doable, but messy.

This region is more easily handled by integrating with respect to x rather than y as you show. Have you learned about the "cylindrical shell" method?

4. Originally Posted by tennisbossen
What is the volume of the solid lying between the curves y= x^2+2x-4 and g=2x-3 …
There are no solids lying between two curves in a plane.

What you're trying to ask is:

What is the volume obtained by rotating, about the y-axis, the region in Quadrant III lying between the curves

f(x) = x^2 + 2x - 4

g(x) = 2x - 3

I agree with Dr. Peterson. Integrating along the x-axis is much easier, using f(x)-g(x) for the height of each shell. This approach requires one integral.

If you want to use disks and integrate along the y-axis, instead, then you'll need to determine a function for the horizontal distance between the curves in terms of y (i.e., the radius of each disk, integrating from y=-5 to y=-4). You'll then need a different function for the horizontal distance between function g and the y-axis (i.e., the radius of each disk, integrating from y=-4 to y=-3). Inverse functions for f and g will help. This approach requires summing two integrals.

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